QUESTION IMAGE
Question
1 - \frac{2}{3}(x - 3) < \frac{1}{3}x + 8
-\frac{2}{3}x + 2 < \frac{1}{3}x + 8
+\frac{2}{3}x +\frac{2}{3}x
2 < x + 8
-8 -8
-6 < x → x > -6
Step1: Start with the inequality
We have the inequality \( 1 - \frac{2}{3}(x - 3) < \frac{1}{3}x + 8 \). First, distribute the \(-\frac{2}{3}\) on the left side: \( 1 - \frac{2}{3}x + 2 < \frac{1}{3}x + 8 \).
Step2: Combine like terms on the left
Combine \(1\) and \(2\) to get \( 3 - \frac{2}{3}x < \frac{1}{3}x + 8 \).
Step3: Add \(\frac{2}{3}x\) to both sides
This gives \( 3 < \frac{1}{3}x + \frac{2}{3}x + 8 \), which simplifies to \( 3 < x + 8 \).
Step4: Subtract 8 from both sides
Subtracting \(8\) from both sides: \( 3 - 8 < x \), so \( -5 < x \)? Wait, but the handwritten work seems to have a different path. Let's re - examine the handwritten steps. The first step in the handwritten work: \( -\frac{2}{3}x+2 < \frac{1}{3}x + 8 \) (after distributing and combining \(1\) and \(2\) as \(3\)? Wait, maybe a miscalculation in the handwritten. Let's do it properly.
Starting over:
Given \( 1-\frac{2}{3}(x - 3)<\frac{1}{3}x + 8 \)
Distribute: \(1-\frac{2}{3}x + 2<\frac{1}{3}x + 8\)
Combine like terms: \(3-\frac{2}{3}x<\frac{1}{3}x + 8\)
Add \(\frac{2}{3}x\) to both sides: \(3<\frac{1}{3}x+\frac{2}{3}x + 8\)
Simplify the right - hand side: \(3 < x + 8\)
Subtract \(8\) from both sides: \(3-8 < x\), so \(x>- 5\). But the handwritten result is \(x > - 6\). There might be an error in the handwritten calculation. Let's check the handwritten steps again.
Looking at the handwritten:
First line: \(1-\frac{2}{3}(x - 3)<\frac{1}{3}x + 8\)
Second line: \(-\frac{2}{3}x + 2<\frac{1}{3}x + 8\) (here, maybe they miscalculated \(1 + 2\) as \(2\) instead of \(3\))
Then add \(\frac{2}{3}x\) to both sides: \(2 Subtract \(8\) from both sides: \(2 - 8
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If we follow the handwritten (erroneous) steps, the solution is \(x > - 6\). If we follow the correct algebraic steps, the solution is \(x>-5\). But based on the handwritten work, the final answer is \(x > - 6\)